HDU 1069 Monkey and Banana(线性DP)
be able to reach the banana by placing one block on the top another to build a tower and climb up to get its favorite food.
of the base and the other dimension was the height.
They want to make sure that the tallest tower possible by stacking blocks can reach the roof. The problem is that, in building a tower, one block could only be placed on top of another block as long as the two base dimensions of the upper block were both strictly
smaller than the corresponding base dimensions of the lower block because there has to be some space for the monkey to step on. This meant, for example, that blocks oriented to have equal-sized bases couldn't be stacked.
The input file will contain one or more test cases. The first line of each test case contains an integer n,
For each test case, print one line containing the case number (they are numbered sequentially starting from 1) and the height of the tallest possible tower in the format "Case case: maximum height = height".
10 20 30
2
6 8 10
5 5 5
7
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
5
31 41 59
26 53 58
97 93 23
84 62 64
33 83 27
0
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342
#include <bits/stdc++.h>
using namespace std;
struct rectangle
{
int x;
int y;
int z;
int size;
};
bool cmp(rectangle a,rectangle b)
{
// if(a.x!=b.x)return a.x>b.x;
// else return a.y>b.y;
return a.size>b.size;
}
vector<rectangle>v;
int n;
//dp[i]:以第i个为顶的最大高度 dp[i]=max(dp[i],dp[k]+h[i])
int dp[]; int main()
{
int i,j;
int cnt=;
while(scanf("%d",&n)&&n)
{
cnt++;
v.clear();
rectangle occ;
v.push_back(occ);
for(i=;i<=n;i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
rectangle temp;
temp.z=x;
temp.x=max(y,z);
temp.y=min(y,z);
temp.size=y*z;
v.push_back(temp);
temp.z=y;
temp.x=max(x,z);
temp.y=min(x,z);
temp.size=x*z;
v.push_back(temp);
temp.z=z;
temp.x=max(x,y);
temp.y=min(x,y);
temp.size=x*y;
v.push_back(temp);
}
sort(v.begin()+,v.end(),cmp);//不要sort占位的
for(i=;i<v.size();i++)//初始化!!假设一个长方体没有任何底座,他自己的高就是最终的高
{
dp[i]=v[i].z;
}
int ans=dp[];
for(i=;i<v.size();i++)
{
for(j=;j<i;j++)//只有i之前的能作为底座
{
if(v[j].x>v[i].x&&v[j].y>v[i].y)
{
dp[i]=max(dp[i],dp[j]+v[i].z); }
ans=max(ans,dp[i]);
}
} printf("Case %d: maximum height = %d\n",cnt,ans);
}
return ;
}
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